1

Property Relationships

Chapter 6

2

dUWQ revrev intint

TdSQ rev int PdVW rev int

dUPdVTdS

Apply the differential form of the first law for a closed

stationary system for an internally reversible process

The T-ds relations :

3

PdvduTds This equation is known as:

First Gibbs equation or First Tds relationshipDivide by T, ..

T

Pdv

T

duds

Although we get this form for internally reversible process, we still can compute s for an irreversible process. This is because S is a point function.

Divide by the mass, you get

4

Second T-ds (Gibbs) relationshipRecall that… Pvuh

vdPPdvdudh Take the differential for both sides

Rearrange to find du

Substitute in the First Tds relationship PdvduTds

vdPPdvdhdu

vdPdhTds Second Tds relationship, or Gibbs equation

5

T

vdP

T

dhds Divide by T, ..

Thus We have two equations for ds

T

vdP

T

dhds

T

Pdv

T

duds

To find s, we have to integrate these equations. Thus we need a relation between du and T (or dh and T).Now we can find entropy change (the LHS of the entropy balance) for liquids and solids

6

2 -Entropy Change of Liquids and Solids

T

Pdv

T

duds

Solids and liquids do not change specific volume appreciably with pressure. That means that dv=0, so the first equation is the easiest to use.

0

T

duds

Thus For solids and liquids

Recall also that For solids and liquids , CdTdu

T

CdT

T

duds

7

1

2lnT

TCs

Integrate to give…

Only true Only true for solids for solids and and liquids!!liquids!!

What if the process is isentropic?

0ln,01

2

T

TCs

The only way this expression can equal 0 is if,

Hence, for solids and liquids, isentropic processes are also isothermal.

21 TT

8

Example(6-7): Effect of Density of a Liquid on Entropy

Liquid methane is commonly used in various cryogenic applications. The critical temperature of methane is 191 K (or -82oC), and thus methane must be maintained below 191 K to keep it in liquid phase. The properties of liquid methane at various temperatures and pressures are given next page.

Determine the entropy change of liquid methane as it undergoes a process from 110 K and 1 MPa to 120 K and 5 MPa

(a) using actual data for methane and

(b) approximating liquid methane as an incompressible substance. What is the error involved in the later case?

9

kgKkjsss /270.0875.4145.522

kgKkjT

TC

T

TCs avg /303.0

110

120ln4785.3lnln

1

2

1

2

10

Example (6-19): Entropy generated when a hot block is dropped in a lake

A 50-kg block of iron casting at 500 K is dropped in a large lake that is at 285 K. The block reaches thermal equilibrium with lake water.

Assuming an average specific heat of 0.45 kJ/kg.K for the iron, determine:

(a) The entropy change of the iron block,

(b) The entropy change of the water lake,

(c) the entropy generated during this process.

11

KkjT

TmCSiron /65.12

500

285ln45.050ln

1

2

(a) The entropy change of the iron block,

(b) The entropy change of the water lake,

we need also to find Q coming out of the system.

UWQ )(0 12 TTmCQ 50 0.45(285 500) 4838Q kJ

KkJT

QS

lakelake /97.16

285

4838

T=500K

Tsurr= 285 K

12

Thus Sg = Stot = Ssys + Slake

T=500K

Tsurr= 285 K

System boundary

(c) the entropy generated during this process.

Sg = Stot = -12.6 + 16.97

= 4.32

Choose the iron block and the lake as the system and treat it is an isolated system.

13

3 -The Entropy Change of Ideal Gases, first relation

The entropy change of an ideal gas can be obtained by substituting du = CvdT and P /T= R/ into Tds relations:

1

2

2

1

12

lnRT

dTTCss v

Tds du pd v

dT dds C R

T

integrating

First relation

du Pdds

T T

14

A second relation for the entropy change of an ideal gas for a process can be obtained by substituting dh = CpdT and /T= R/P into Tds relations:

2

1 1

212 P

PlnR

T

dTTCss p

Tds dh vdp p

dT dpds C R

T p

integrating

Second relation

dh vdpds

T T

15

1

22

1

12 ln

RT

dTTCss v

2

1 1

212 P

PlnR

T

dTTCss p

The integration of the first term on the RHS can be done via two methods:

1. Assume constant Cp and constant Cv (Approximate Analysis)

2. Evaluate these integrals exactly and tabulate the data (Exact Analysis)

16

Method 1: Constant specific heats (Approximate Analysis)First relation

1

2

1

2 lnlnv

vR

T

TCs v

Only true for ideal gases, assuming constant heat capacitiesSecond relation

Only true for ideal gases, assuming constant heat capacities

1

2

1

2 lnlnP

PR

T

TCs p

1

22

1

12 ln

RT

dTTCss v

2

1 1

212 ln

P

PR

T

dTTCss p

17

Sometimes it is more convenient to calculate the change in entropy per mole, instead of per unit mass

1

2

1

212 lnln

v

vR

T

TCsss uv

1

2

1

212 lnln

P

PR

T

TCsss up

kJ/kmol. K

kJ/kmol. K

Ru is the universal gas constant

18

Method 2: Variable specific heats (Exact Analysis)

We could substitute in the equations for Cv and Cp, and perform the integrations Cp = a + bT + cT2 + dT3

But this is time consuming. Someone already did the integrations and

tabulated them for us (table A-17) They assume absolute 0 as the starting point

1

22

1 P

PlnR

T

dTCs p We use the

second relation

191

20

1

0

2 P

PlnRsss

The integral is expressed as:

2

1

)(T

T p T

dTTC

0

1

0

2

2

1

ssT

dT)T(C

T

Tp

T

p T

dT)T(Cs

0

0

Where

is tabulated in Table A-17

Therefore

K.kg/kJ:unit

2

0)(

T

p T

dTTC 1

0)(

T

p T

dTTC

20

From this equation, It can be seen that the entropy of an ideal gas is not a function only of the temperature ( as was the internal energy) but also of the pressure or the specific volume.

The function s° represents only the temperature-dependent part of entropy

1

20

1

0

2 P

PlnRsss

Temperature dependencePressure

dependence

Is s = f (T) only? like u for an ideal gas. Let us see

21

How about the other relation

1

22

1 v

vlnR

T

dTCs v

We can develop another relation for the entropy changed based on the above relation

but this will require the definition of another function and tabulating it which is not practical.

T

v T

dT)T(C?

0

22

6-4 Isentropic Processes The entropy of a fixed mass can be

changed by

1. Heat transfer,

2. Irreversibilities It follows that the entropy of a system

will not change if we have

1. Adiabatic process,

2. Internally reversible process. Therefore, we define the following:

23

Isentropic Processes of Ideal Gases

Many real processes can be modeled as isentropic

Isentropic processes are the standard against which we should measure efficiency

We need to develop isentropic relationships for ideal gases, just like we developed for solids and liquids

24

1

2

1

2 lnlnv

vR

T

TCs v

For the isentropic case, S=0. Thus

1

2

1

2 lnlnv

vR

T

TCv

Constant specific heats (1st relation)

vC

R

v v

v

v

v

C

R

T

T

2

1

1

2

1

2 lnlnln

Recall

Recall also from ch 2, the following relations..…

11 kC/CC/RCCR vpvvp

1

2

1

1

2

k

v

v

T

T Only applies to ideal gases, with constant specific heats

25

0lnln1

2

1

2

P

PR

T

TCs p

pC

R

p P

P

P

P

C

R

T

T

1

2

1

2

1

2 lnlnln

k

k

P

P

T

T1

1

2

1

2

Only applies to ideal gases, with constant specific heats

Constant specific heats (2nd relation)

Recall..… or1/ kCR v

1/ p

kR C

k

26

Since…

k

k

P

P

T

T1

1

2

1

2

1

2

1

1

2

k

v

v

T

Tand

k

kk

P

P

v

v1

1

2

1

2

1

Which can be simplified to…

1

2

2

1

P

P

v

vk

Third isentropic relationship

HENCE

27

1

2

1

1

2

k

v

v

T

T k

k

P

P

T

T1

1

2

1

2

1

2

2

1

P

P

v

vk

constantTvk 1

Compact form

constantTP k

k

1

constantPvk

Full form of Isentropic relations of Ideal Gases

Valid for only for 1- Ideal gas 2- Isentropic process 3- Constant specific heats

28

That works if the specific heat constants can be approximated as constant, but what if

that’s not a good assumption?

1

201

02 ln

P

PRsss

We need to use the exact treatment 0

1

201

02 ln

P

PRss

This equation is a good way to evaluate property changes, but it can be tedious if you know the volume ratio instead of the pressure ratio

29

1

201

02 ln

P

PRss

R

ss

P

P 01

02

1

2 exp

02

20

1 1

exp

exp

sRP

P sR

s20 is only a

function of temperature!!!

1

201

02 ln

P

P

R

ss Rename the exponential term as Pr , (relative pressure) which is only a function of temperature, and is tabulated on the ideal gas tables

02

20

11

exp

exp

r

r

sR P

PsR

2 2

1 1

r

r

P P

P P

30

1

2

1

2

r

r

P

P

P

P

You can use either of the following 2 equations

1

201

02 ln

P

PRss

This is good if you know the pressure ratio but how about if you know only the volume ratioIn this case, we use the ideal gas law

2

22

1

11

T

vP

T

vP

2

1

1

2

1

2

P

P

T

T

v

v

2

1

1

2

r

r

P

P

T

T

1

1

2

2

T

P

P

T r

r

1

2

r

r

v

v

where

rr P/Tv

Remember, these relationships only hold for ideal gases and isentropic processes

1

2

1

2

r

r

v

v

v

v

31

Example (6-10):

Isentropic Compression of Air in a Car Engine

Air is compressed in a car engine from 22oC and 95 kPa in a reversible and adiabatic manner. If the compression ratio V1/V2 of this piston-cylinder device is 8, determine the final temperature of the air. <Answer: 662.7 K>

Sol: